3.16 \(\int (c+d x) (b \tanh (e+f x))^{5/2} \, dx\)

Optimal. Leaf size=1392 \[ -\frac {d \tanh ^{-1}\left (\frac {\sqrt {b \tanh (e+f x)}}{\sqrt {-b}}\right )^2 (-b)^{5/2}}{2 f^2}-\frac {(c+d x) \tanh ^{-1}\left (\frac {\sqrt {b \tanh (e+f x)}}{\sqrt {-b}}\right ) (-b)^{5/2}}{f}+\frac {d \tanh ^{-1}\left (\frac {\sqrt {b \tanh (e+f x)}}{\sqrt {-b}}\right ) \log \left (\frac {2}{1-\frac {\sqrt {b \tanh (e+f x)}}{\sqrt {-b}}}\right ) (-b)^{5/2}}{f^2}-\frac {d \tanh ^{-1}\left (\frac {\sqrt {b \tanh (e+f x)}}{\sqrt {-b}}\right ) \log \left (\frac {2 \left (\sqrt {b}-\sqrt {b \tanh (e+f x)}\right )}{\left (\sqrt {-b}+\sqrt {b}\right ) \left (1-\frac {\sqrt {b \tanh (e+f x)}}{\sqrt {-b}}\right )}\right ) (-b)^{5/2}}{2 f^2}-\frac {d \tanh ^{-1}\left (\frac {\sqrt {b \tanh (e+f x)}}{\sqrt {-b}}\right ) \log \left (-\frac {2 \left (\sqrt {b}+\sqrt {b \tanh (e+f x)}\right )}{\left (\sqrt {-b}-\sqrt {b}\right ) \left (1-\frac {\sqrt {b \tanh (e+f x)}}{\sqrt {-b}}\right )}\right ) (-b)^{5/2}}{2 f^2}-\frac {d \tanh ^{-1}\left (\frac {\sqrt {b \tanh (e+f x)}}{\sqrt {-b}}\right ) \log \left (\frac {2}{\frac {\sqrt {b \tanh (e+f x)}}{\sqrt {-b}}+1}\right ) (-b)^{5/2}}{f^2}+\frac {d \text {Li}_2\left (1-\frac {2}{1-\frac {\sqrt {b \tanh (e+f x)}}{\sqrt {-b}}}\right ) (-b)^{5/2}}{2 f^2}-\frac {d \text {Li}_2\left (1-\frac {2 \left (\sqrt {b}-\sqrt {b \tanh (e+f x)}\right )}{\left (\sqrt {-b}+\sqrt {b}\right ) \left (1-\frac {\sqrt {b \tanh (e+f x)}}{\sqrt {-b}}\right )}\right ) (-b)^{5/2}}{4 f^2}-\frac {d \text {Li}_2\left (\frac {2 \left (\sqrt {b}+\sqrt {b \tanh (e+f x)}\right )}{\left (\sqrt {-b}-\sqrt {b}\right ) \left (1-\frac {\sqrt {b \tanh (e+f x)}}{\sqrt {-b}}\right )}+1\right ) (-b)^{5/2}}{4 f^2}+\frac {d \text {Li}_2\left (1-\frac {2}{\frac {\sqrt {b \tanh (e+f x)}}{\sqrt {-b}}+1}\right ) (-b)^{5/2}}{2 f^2}+\frac {b^{5/2} d \tanh ^{-1}\left (\frac {\sqrt {b \tanh (e+f x)}}{\sqrt {b}}\right )^2}{2 f^2}-\frac {2 b (c+d x) (b \tanh (e+f x))^{3/2}}{3 f}+\frac {2 b^{5/2} d \tan ^{-1}\left (\frac {\sqrt {b \tanh (e+f x)}}{\sqrt {b}}\right )}{3 f^2}+\frac {b^{5/2} (c+d x) \tanh ^{-1}\left (\frac {\sqrt {b \tanh (e+f x)}}{\sqrt {b}}\right )}{f}+\frac {2 b^{5/2} d \tanh ^{-1}\left (\frac {\sqrt {b \tanh (e+f x)}}{\sqrt {b}}\right )}{3 f^2}-\frac {b^{5/2} d \tanh ^{-1}\left (\frac {\sqrt {b \tanh (e+f x)}}{\sqrt {b}}\right ) \log \left (\frac {2 \sqrt {b}}{\sqrt {b}-\sqrt {b \tanh (e+f x)}}\right )}{f^2}+\frac {b^{5/2} d \tanh ^{-1}\left (\frac {\sqrt {b \tanh (e+f x)}}{\sqrt {b}}\right ) \log \left (\frac {2 \sqrt {b}}{\sqrt {b}+\sqrt {b \tanh (e+f x)}}\right )}{f^2}-\frac {b^{5/2} d \tanh ^{-1}\left (\frac {\sqrt {b \tanh (e+f x)}}{\sqrt {b}}\right ) \log \left (\frac {2 \sqrt {b} \left (\sqrt {-b}-\sqrt {b \tanh (e+f x)}\right )}{\left (\sqrt {-b}-\sqrt {b}\right ) \left (\sqrt {b}+\sqrt {b \tanh (e+f x)}\right )}\right )}{2 f^2}-\frac {b^{5/2} d \tanh ^{-1}\left (\frac {\sqrt {b \tanh (e+f x)}}{\sqrt {b}}\right ) \log \left (\frac {2 \sqrt {b} \left (\sqrt {-b}+\sqrt {b \tanh (e+f x)}\right )}{\left (\sqrt {-b}+\sqrt {b}\right ) \left (\sqrt {b}+\sqrt {b \tanh (e+f x)}\right )}\right )}{2 f^2}-\frac {b^{5/2} d \text {Li}_2\left (1-\frac {2 \sqrt {b}}{\sqrt {b}-\sqrt {b \tanh (e+f x)}}\right )}{2 f^2}-\frac {b^{5/2} d \text {Li}_2\left (1-\frac {2 \sqrt {b}}{\sqrt {b}+\sqrt {b \tanh (e+f x)}}\right )}{2 f^2}+\frac {b^{5/2} d \text {Li}_2\left (1-\frac {2 \sqrt {b} \left (\sqrt {-b}-\sqrt {b \tanh (e+f x)}\right )}{\left (\sqrt {-b}-\sqrt {b}\right ) \left (\sqrt {b}+\sqrt {b \tanh (e+f x)}\right )}\right )}{4 f^2}+\frac {b^{5/2} d \text {Li}_2\left (1-\frac {2 \sqrt {b} \left (\sqrt {-b}+\sqrt {b \tanh (e+f x)}\right )}{\left (\sqrt {-b}+\sqrt {b}\right ) \left (\sqrt {b}+\sqrt {b \tanh (e+f x)}\right )}\right )}{4 f^2}-\frac {4 b^2 d \sqrt {b \tanh (e+f x)}}{3 f^2} \]

[Out]

2/3*b^(5/2)*d*arctan((b*tanh(f*x+e))^(1/2)/b^(1/2))/f^2-(-b)^(5/2)*(d*x+c)*arctanh((b*tanh(f*x+e))^(1/2)/(-b)^
(1/2))/f-1/2*(-b)^(5/2)*d*arctanh((b*tanh(f*x+e))^(1/2)/(-b)^(1/2))^2/f^2+2/3*b^(5/2)*d*arctanh((b*tanh(f*x+e)
)^(1/2)/b^(1/2))/f^2+b^(5/2)*(d*x+c)*arctanh((b*tanh(f*x+e))^(1/2)/b^(1/2))/f+1/2*b^(5/2)*d*arctanh((b*tanh(f*
x+e))^(1/2)/b^(1/2))^2/f^2-b^(5/2)*d*arctanh((b*tanh(f*x+e))^(1/2)/b^(1/2))*ln(2*b^(1/2)/(b^(1/2)-(b*tanh(f*x+
e))^(1/2)))/f^2+b^(5/2)*d*arctanh((b*tanh(f*x+e))^(1/2)/b^(1/2))*ln(2*b^(1/2)/(b^(1/2)+(b*tanh(f*x+e))^(1/2)))
/f^2-1/2*b^(5/2)*d*arctanh((b*tanh(f*x+e))^(1/2)/b^(1/2))*ln(2*b^(1/2)*((-b)^(1/2)-(b*tanh(f*x+e))^(1/2))/((-b
)^(1/2)-b^(1/2))/(b^(1/2)+(b*tanh(f*x+e))^(1/2)))/f^2-1/2*b^(5/2)*d*arctanh((b*tanh(f*x+e))^(1/2)/b^(1/2))*ln(
2*b^(1/2)*((-b)^(1/2)+(b*tanh(f*x+e))^(1/2))/((-b)^(1/2)+b^(1/2))/(b^(1/2)+(b*tanh(f*x+e))^(1/2)))/f^2+(-b)^(5
/2)*d*arctanh((b*tanh(f*x+e))^(1/2)/(-b)^(1/2))*ln(2/(1-(b*tanh(f*x+e))^(1/2)/(-b)^(1/2)))/f^2-1/2*(-b)^(5/2)*
d*arctanh((b*tanh(f*x+e))^(1/2)/(-b)^(1/2))*ln(2*(b^(1/2)-(b*tanh(f*x+e))^(1/2))/((-b)^(1/2)+b^(1/2))/(1-(b*ta
nh(f*x+e))^(1/2)/(-b)^(1/2)))/f^2-1/2*(-b)^(5/2)*d*arctanh((b*tanh(f*x+e))^(1/2)/(-b)^(1/2))*ln(-2*(b^(1/2)+(b
*tanh(f*x+e))^(1/2))/((-b)^(1/2)-b^(1/2))/(1-(b*tanh(f*x+e))^(1/2)/(-b)^(1/2)))/f^2-(-b)^(5/2)*d*arctanh((b*ta
nh(f*x+e))^(1/2)/(-b)^(1/2))*ln(2/(1+(b*tanh(f*x+e))^(1/2)/(-b)^(1/2)))/f^2-1/2*b^(5/2)*d*polylog(2,1-2*b^(1/2
)/(b^(1/2)-(b*tanh(f*x+e))^(1/2)))/f^2-1/2*b^(5/2)*d*polylog(2,1-2*b^(1/2)/(b^(1/2)+(b*tanh(f*x+e))^(1/2)))/f^
2+1/4*b^(5/2)*d*polylog(2,1-2*b^(1/2)*((-b)^(1/2)-(b*tanh(f*x+e))^(1/2))/((-b)^(1/2)-b^(1/2))/(b^(1/2)+(b*tanh
(f*x+e))^(1/2)))/f^2+1/4*b^(5/2)*d*polylog(2,1-2*b^(1/2)*((-b)^(1/2)+(b*tanh(f*x+e))^(1/2))/((-b)^(1/2)+b^(1/2
))/(b^(1/2)+(b*tanh(f*x+e))^(1/2)))/f^2+1/2*(-b)^(5/2)*d*polylog(2,1-2/(1-(b*tanh(f*x+e))^(1/2)/(-b)^(1/2)))/f
^2-1/4*(-b)^(5/2)*d*polylog(2,1-2*(b^(1/2)-(b*tanh(f*x+e))^(1/2))/((-b)^(1/2)+b^(1/2))/(1-(b*tanh(f*x+e))^(1/2
)/(-b)^(1/2)))/f^2-1/4*(-b)^(5/2)*d*polylog(2,1+2*(b^(1/2)+(b*tanh(f*x+e))^(1/2))/((-b)^(1/2)-b^(1/2))/(1-(b*t
anh(f*x+e))^(1/2)/(-b)^(1/2)))/f^2+1/2*(-b)^(5/2)*d*polylog(2,1-2/(1+(b*tanh(f*x+e))^(1/2)/(-b)^(1/2)))/f^2-4/
3*b^2*d*(b*tanh(f*x+e))^(1/2)/f^2-2/3*b*(d*x+c)*(b*tanh(f*x+e))^(3/2)/f

________________________________________________________________________________________

Rubi [F]  time = 0.13, antiderivative size = 0, normalized size of antiderivative = 0.00, number of steps used = 0, number of rules used = 0, integrand size = 0, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.000, Rules used = {} \[ \int (c+d x) (b \tanh (e+f x))^{5/2} \, dx \]

Verification is Not applicable to the result.

[In]

Int[(c + d*x)*(b*Tanh[e + f*x])^(5/2),x]

[Out]

(2*b^(5/2)*d*ArcTan[Sqrt[b*Tanh[e + f*x]]/Sqrt[b]])/(3*f^2) + (2*b^(5/2)*d*ArcTanh[Sqrt[b*Tanh[e + f*x]]/Sqrt[
b]])/(3*f^2) - (4*b^2*d*Sqrt[b*Tanh[e + f*x]])/(3*f^2) - (2*b*(c + d*x)*(b*Tanh[e + f*x])^(3/2))/(3*f) + b^2*D
efer[Int][(c + d*x)*Sqrt[b*Tanh[e + f*x]], x]

Rubi steps

\begin {align*} \int (c+d x) (b \tanh (e+f x))^{5/2} \, dx &=-\frac {2 b (c+d x) (b \tanh (e+f x))^{3/2}}{3 f}+b^2 \int (c+d x) \sqrt {b \tanh (e+f x)} \, dx+\frac {(2 b d) \int (b \tanh (e+f x))^{3/2} \, dx}{3 f}\\ &=-\frac {4 b^2 d \sqrt {b \tanh (e+f x)}}{3 f^2}-\frac {2 b (c+d x) (b \tanh (e+f x))^{3/2}}{3 f}+b^2 \int (c+d x) \sqrt {b \tanh (e+f x)} \, dx+\frac {\left (2 b^3 d\right ) \int \frac {1}{\sqrt {b \tanh (e+f x)}} \, dx}{3 f}\\ &=-\frac {4 b^2 d \sqrt {b \tanh (e+f x)}}{3 f^2}-\frac {2 b (c+d x) (b \tanh (e+f x))^{3/2}}{3 f}+b^2 \int (c+d x) \sqrt {b \tanh (e+f x)} \, dx-\frac {\left (2 b^4 d\right ) \operatorname {Subst}\left (\int \frac {1}{\sqrt {x} \left (-b^2+x^2\right )} \, dx,x,b \tanh (e+f x)\right )}{3 f^2}\\ &=-\frac {4 b^2 d \sqrt {b \tanh (e+f x)}}{3 f^2}-\frac {2 b (c+d x) (b \tanh (e+f x))^{3/2}}{3 f}+b^2 \int (c+d x) \sqrt {b \tanh (e+f x)} \, dx-\frac {\left (4 b^4 d\right ) \operatorname {Subst}\left (\int \frac {1}{-b^2+x^4} \, dx,x,\sqrt {b \tanh (e+f x)}\right )}{3 f^2}\\ &=-\frac {4 b^2 d \sqrt {b \tanh (e+f x)}}{3 f^2}-\frac {2 b (c+d x) (b \tanh (e+f x))^{3/2}}{3 f}+b^2 \int (c+d x) \sqrt {b \tanh (e+f x)} \, dx+\frac {\left (2 b^3 d\right ) \operatorname {Subst}\left (\int \frac {1}{b-x^2} \, dx,x,\sqrt {b \tanh (e+f x)}\right )}{3 f^2}+\frac {\left (2 b^3 d\right ) \operatorname {Subst}\left (\int \frac {1}{b+x^2} \, dx,x,\sqrt {b \tanh (e+f x)}\right )}{3 f^2}\\ &=\frac {2 b^{5/2} d \tan ^{-1}\left (\frac {\sqrt {b \tanh (e+f x)}}{\sqrt {b}}\right )}{3 f^2}+\frac {2 b^{5/2} d \tanh ^{-1}\left (\frac {\sqrt {b \tanh (e+f x)}}{\sqrt {b}}\right )}{3 f^2}-\frac {4 b^2 d \sqrt {b \tanh (e+f x)}}{3 f^2}-\frac {2 b (c+d x) (b \tanh (e+f x))^{3/2}}{3 f}+b^2 \int (c+d x) \sqrt {b \tanh (e+f x)} \, dx\\ \end {align*}

________________________________________________________________________________________

Mathematica [F]  time = 39.85, size = 0, normalized size = 0.00 \[ \int (c+d x) (b \tanh (e+f x))^{5/2} \, dx \]

Verification is Not applicable to the result.

[In]

Integrate[(c + d*x)*(b*Tanh[e + f*x])^(5/2),x]

[Out]

Integrate[(c + d*x)*(b*Tanh[e + f*x])^(5/2), x]

________________________________________________________________________________________

fricas [F(-2)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Exception raised: TypeError} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((d*x+c)*(b*tanh(f*x+e))^(5/2),x, algorithm="fricas")

[Out]

Exception raised: TypeError >>  Error detected within library code:   integrate: implementation incomplete (co
nstant residues)

________________________________________________________________________________________

giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int {\left (d x + c\right )} \left (b \tanh \left (f x + e\right )\right )^{\frac {5}{2}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((d*x+c)*(b*tanh(f*x+e))^(5/2),x, algorithm="giac")

[Out]

integrate((d*x + c)*(b*tanh(f*x + e))^(5/2), x)

________________________________________________________________________________________

maple [F]  time = 0.22, size = 0, normalized size = 0.00 \[ \int \left (d x +c \right ) \left (b \tanh \left (f x +e \right )\right )^{\frac {5}{2}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((d*x+c)*(b*tanh(f*x+e))^(5/2),x)

[Out]

int((d*x+c)*(b*tanh(f*x+e))^(5/2),x)

________________________________________________________________________________________

maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int {\left (d x + c\right )} \left (b \tanh \left (f x + e\right )\right )^{\frac {5}{2}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((d*x+c)*(b*tanh(f*x+e))^(5/2),x, algorithm="maxima")

[Out]

integrate((d*x + c)*(b*tanh(f*x + e))^(5/2), x)

________________________________________________________________________________________

mupad [F]  time = 0.00, size = -1, normalized size = -0.00 \[ \int {\left (b\,\mathrm {tanh}\left (e+f\,x\right )\right )}^{5/2}\,\left (c+d\,x\right ) \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((b*tanh(e + f*x))^(5/2)*(c + d*x),x)

[Out]

int((b*tanh(e + f*x))^(5/2)*(c + d*x), x)

________________________________________________________________________________________

sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \left (b \tanh {\left (e + f x \right )}\right )^{\frac {5}{2}} \left (c + d x\right )\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((d*x+c)*(b*tanh(f*x+e))**(5/2),x)

[Out]

Integral((b*tanh(e + f*x))**(5/2)*(c + d*x), x)

________________________________________________________________________________________